We can solve this problem by first writing the chemical equation for the reaction:
#"X" + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)#
We know that the temperature and pressure remained unchanged throughout the reaction, so let's let these values be those at STP to make things much easier.
At STP, one mole of an (ideal) gas has a volume of #22.41# #"L"#. Let's convert all 4 of our given volumes to moles, for reference:

#"X"#: #10cancel("cm"^3)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1color(white)(l)"L")/(1cancel("dm"^3))) = 0.010# #"L X"#

#"original O"_2#: #70cancel("cm"^3)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1color(white)(l)"L")/(1cancel("dm"^3))) = 0.070# #"L O"_2#

#"CO"_2#: #30cancel("cm"^3)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1color(white)(l)"L")/(1cancel("dm"^3))) = 0.030# #"L CO"_2#

#"final O"_2#: #20cancel("cm"^3)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1color(white)(l)"L")/(1cancel("dm"^3))) = 0.020# #"L O"_2#
Now, using the fact that #1# #"mol = 22.41"# #"L"#, let's convert all of them to #"mol"#:

#"X"#: #4.46xx10^4# #"mol"#

#"original O"_2#: #0.00312# #"mol"#

#"CO"_2#: #0.00134# #"mol"#

#"final O"_2#: #8.92xx10^4# #"mol"#
The number of carbon atoms in #"X"# equals the ratio of moles of #"CO"_2# to moles of #"X"# (the coefficient of #"CO"_2# must equal the number of atoms of #"C"# in compound #"X"#). Therefore,
#"atoms C" = ("mol CO"_2)/("mol X") = (color(green)(0.00134"mol CO"_2))/(color(red)(4.46xx10^4"mol X")) = color(blue)(3.0#
There are thus #3# atoms of carbon in compound #"X"#, so the answer now is either (b) or (c).
We now have to find the number of #"H"# atoms in #"X"#. We can do this by asking, "which compound would use up #1.12# moles of #"O"_2#?"
(The #1.12# moles value was found by subtracting the final moles of oxygen from the initial value.)
Let's write what the equation looks like so far:
#"C"_ (color(blue)(3))"H"_x + "O"_2(g) rarr color(blue)(3)"CO"_2(g) + "H"_2"O"(g)#
Similarly to how we found the number of #"C"# atoms in the hydrocarbon, let's find the ratio of oxygen used to hydrocarbon used to find the coefficient of #"O"_2#:
#"coefficient O"_2 = ("mol O"_2)/("mol X") = (color(green)(1.12"mol O"_2))/(color(red)(0.224"mol X")) = color(orange)(5.0#
The coefficient of #"O"_2# is thus #color(orange)(5#, so our equation is now
#"C"_ (color(blue)(3))"H"_x + color(orange)(5)"O"_2(g) rarr color(blue)(3)"CO"_2(g) + "H"_2"O"(g)#
According to the law of conservation of mass, the number of atoms of #"O"# must be equal on both sides of the equation. What we have so far is

Left: #10# #"atoms O"#

Right: #7# #"atoms O"#
In order to balance it, we can simply place a #color(purple)(4# in front of the #"H"_2"O"#:
#"C"_ (color(blue)(3))"H"_x + color(orange)(5)"O"_2(g) rarr color(blue)(3)"CO"_2(g) + color(purple)(4)"H"_2"O"(g)#
One final step! Again, the law of conservation of mass says the atoms of #"H"# must be equal on both sides, so the number of hydrogen atoms in the hydrocarbon must be 8 to balance it.
And therefore, the hydrocarbon is propane, #"C"_3"H"_8#, and thus (c) is the correct answer.