Answer
$ 25 ^\circ \rm C$
Work Step by Step
Let's assume that the system is perfectly isolated, so the heat loss from the one-mole monatomic gas is gained by the one-mole solid element.
Hence,
$$(\Delta E_{th})_{gas}+(\Delta E_{th})_{solid}=0 $$
$$(\Delta E_{th})_{gas}=-(\Delta E_{th})_{solid}$$
$$n_1C_{\rm v,1}\Delta T_1=-n_2C_{\rm v,2}\Delta T_2$$
where $1$ refers to the gas and $2$ refers to the solid.
Hence,
$$\Delta T_2=-\dfrac{n_1C_{\rm v,1}\Delta T_1}{n_2C_{\rm v,2}}$$
Plugging the known and noting that $\Delta T_1=-50^\circ$ C since the temperature of the gas decreases.
$$\Delta T_2=-\dfrac{(1)(12.5)(-50)}{(1) (25)}=\color{red}{\bf 25}\;\rm K=\color{red}{\bf 25}^\circ C$$
