Answer
$C_p=34.4\frac{J}{mol.K}$
Work Step by Step
We know that;
$C_p=\frac{Q}{n\Delta T}$.......eq(1)
But as $\Delta T=\frac{p\Delta V}{nR}$
Thus, eq(1) becomes
$C_p=\frac{Q}{n\frac{p\Delta V}{nR}}$
$C_p=\frac{RQ}{p \Delta V}$
We plug in the known values to obtain:
$C_p=\frac{8.31(20.9)}{1.01\times 10^5\times50\times 10^{-6}}=34.4\frac{J}{mol.K}$
