Here's what I got.
The idea here is that because the pressure and the number of moles of gas are being kept constant, the volume of the gas will have a direct relationship to its temperature, as described by Charles' Law.
Simply put, when the temperature of the increases, the volume of the gas increases by the same factor. Similarly, when the temperature of the gas decreases, the volume decreases by the same factor.
Mathematically, this can be written as
#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#
#V_1#and #T_1#represent the volume and the temperature of the gas at an initial state
#V_2#and #T_2#represent the volume and the temperature of the gas at a final state
Before moving on, make sure that you convert the temperature from degrees Celsius to Kelvin
#T = 60^@"C" + 273.15 = "333.15 K"#
Now, you know that the volume of the gas must triple as a result of the increase in temperature. Even without setting up the equation, you should be able to say that in order for that to happen, the temperature must triple as well.
#T_2 = 3 * T#
#T_2 = 3 * "333.15 K" = "999.45 K"#
You can show that this is the case by using the equation. You know that
#V_2 = 3 * V_1#
Rearrange the equation to solve for
#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#
Plug in your values to find
#T_2 = (3 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * "333.15 K" = "999.45 K"#
Rounded to one significant figure, the number of sig figs you have for the initial temperature of the gas, the answer will be
#color(darkgreen)(ul(color(black)(T_2 = "1000 K")))#
If you want, you can express this in degrees Celsius
#t_2[""^@"C"] = "999.45 K" - 273.15 = 726.3^@"C"#
Ince again, you must round the answer to one significant figure
#color(darkgreen)(ul(color(black)(t_2 = 700^@"C")))#