All you have to do here is to use the fact that pressure and volume have an inverse relationship when temperature and number of moles are kept constant
In other words, increasing the pressure will cause the volume to decrease. Similarly, decreasing the pressure will cause the volume to increase.
In your case, the pressure is increasing
#"242 torr " -> " 1210 torr"#
so right from the start, you should expect the volume of the gas to be
Mathematically, you will have
#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#
#P_1#and #V_1#represent the pressure and volume of the gas at an initial state
#P_2#and #V_2#represent the pressure and volume of the gas at a final state
Rearrange to solve for
#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#
Plug in your values to find
#V_2 = (242 color(red)(cancel(color(black)("torr"))))/(1210color(red)(cancel(color(black)("torr")))) * "17.0 L" = color(darkgreen)(ul(color(black)("3.40 L")))#
The answer is rounded to three sig figs.
As predicted, the volume decreased as a result of the increase in pressure.