Answer: 34.482 atm

The external pressure which must be applied on solution side to stop the process of osmosis is called osmotic pressure of the solution. It is calculated as $\pi = \Sigma \;i_nC_nRT$

In this case $\pi_{NaCl} = 0.1\; \times 2 RT$, $\pi_{Na_2SO_4} = 0.2 \times 3 RT$ and $\pi_{HA} = 0.5\; RT \times 1.2$

$R$ is the ideal gas constant $ = 0.082 \; $$\text{lit.atm.mol}^{-1}$$ \text{K}^{-1}$

$\Rightarrow \pi = 0.0821 \times 300 (0.2 + 0.6 + 0.6) = 34.482\; atm$