One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\dfrac{5}{3}$, the final temperature of gas will be :-

In an adiabatic process no heat is exchanged: $Q = 0$, so by the first law $0 = \Delta U + W \Rightarrow \Delta U = -W$.
The gas does work $W = 6R$, so $\Delta U = -6R$.
For an ideal gas, $\Delta U = \mu C_V \Delta T$, with $C_V = \dfrac{R}{\gamma - 1}$.
Here $\gamma = \dfrac{5}{3}$, so $C_V = \dfrac{R}{\frac{5}{3} - 1} = \dfrac{3}{2}R$
Therefore $-6R = 1 \times \dfrac{3}{2}R\,\Delta T$
$\Rightarrow \Delta T = -4$ K
$T_{final} = (T - 4)$ K (the gas cools as it does work at the cost of its internal energy).