Graph between  $\log \left(\frac{x}{m}\right)$ and  $\text{log P}$ is a straight line at angle  ${45}^0$ with intercept OA as shown. Hence,  $\left(\frac{x}{m}\right)$ at a pressure of 2 atm  is

Applying freudlich adsorption isotherm, 
$\frac{x}{m}=k{p}^{1/n}$  
Where,  ‘x’ is the mass of the gas adsorbed on a mass  ‘m’ of the adsorbent at a pressure  P.k and  n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
$\begin{array}{l}\Rightarrow \log x/m=\log k+\frac{1}{n}\log p\\ k=2,\frac{1}{n}=1\\ \Rightarrow \frac{x}{m}=kp\\ \therefore \frac{x}{m}=4\end{array}$