At STP, 22.4 L of any gas is 1 mole.
$$
\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \text { moles }=n
$$

In adiabatic process,
$$
\begin{array}{rlrl}
& T V^{\gamma-1} & =\text { constant } \\
& \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\
& \text { or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
\end{array}
$$

$$
\begin{aligned}
& \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text { for monoatomic He gas. } \\
& \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\
& \because \text { Further in adiabatic process, } \\
& Q=0 \\
& \therefore \quad W+\Delta U=0 \\
& \text { or } \quad W=-\Delta U \\
& =-n C_V \Delta T \\
& =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\
& =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\
& =-\frac{9}{8} R T_1 \\
&
\end{aligned}
$$

$\therefore$ Correct option is (a).
Analysis of Question
(i) From calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.