Step-by-step Solution:

1. Identify Given Information and Process:

   • The gas is Helium (He), which is a monoatomic ideal gas.
   • Initial volume, $V_1 = 5.6$ liters.
   • Initial state is at Standard Temperature and Pressure (STP). The initial temperature is denoted as $T_1$.
   • Final volume, $V_2 = 0.7$ liters.
   • The process is an adiabatic compression, which means there is no heat exchange with the surroundings ($Q=0$).

2. Calculate the Number of Moles (n):

   • At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters.
   • The given initial volume is 5.6 liters. Therefore, the number of moles of helium gas is: $n = \frac{\text{Volume at STP}}{\text{Molar volume at STP}} = \frac{5.6 \text{ L}}{22.4 \text{ L/mol}} = \frac{1}{4} \text{ mol}$

3. Determine the Adiabatic Index (γ) for Helium:

   • Helium is a monoatomic gas. For a monoatomic gas, the number of degrees of freedom is $f=3$.
   • The molar specific heat at constant volume is $C_V = \frac{f}{2}R = \frac{3}{2}R$.
   • The molar specific heat at constant pressure is $C_P = C_V + R = \frac{3}{2}R + R = \frac{5}{2}R$.
   • The adiabatic index, $\gamma$, is the ratio of specific heats: $\gamma = \frac{C_P}{C_V} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$

4. Calculate the Final Temperature (T₂):

   • For an adiabatic process, the relationship between temperature (T) and volume (V) is given by the equation $TV^{\gamma-1} = \text{constant}$.
   • Therefore, for the initial and final states, we have $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$.
   • Solving for the final temperature, $T_2$: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$
   • We are given $V_1 = 5.6$ L and $V_2 = 0.7$ L. The ratio of volumes is $\frac{V_1}{V_2} = \frac{5.6}{0.7} = 8$.
   • The exponent is $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
   • Substituting these values into the equation for $T_2$: $T_2 = T_1 (8)^{2/3} = T_1 ((2^3))^{2/3} = T_1 (2^{3 \times 2/3}) = T_1 (2^2) = 4T_1$

5. Calculate the Work Done in the Process:

   • The question asks for the work done. For an adiabatic process, the work done by the gas is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
   • Since the gas is compressed ($V_2 < V_1$), work is done on the gas, and the work done by the gas will be negative. The positive options suggest we need to find the magnitude of the work done, which is equivalent to the work done on the gas.
   • From the First Law of Thermodynamics, $\Delta U = Q - W$. For an adiabatic process, $Q=0$, so $\Delta U = -W$. This means the work done on the gas ($W_{on} = -W$) is equal to the change in internal energy ($\Delta U$).
   • The change in internal energy is given by $\Delta U = nC_V(T_2 - T_1)$.
   • Therefore, the work done is: $W_{\text{on}} = \Delta U = nC_V(T_2 - T_1)$
   • Substituting the values we found:
     • $n = 1/4$
     • $C_V = \frac{3}{2}R$
     • $T_2 = 4T_1$ $W = \left(\frac{1}{4}\right) \left(\frac{3}{2}R\right) (4T_1 - T_1) = \left(\frac{3}{8}R\right)(3T_1) = \frac{9}{8}RT_1$

6. Conclusion: The work done in the process is $\frac{9}{8}RT_1$. This corresponds to option A.