According to Freundlich Isotherm, the mass of gas adsorbed can be calculated as,

$\frac{\mathrm{x}}{\mathrm{m}}={\mathrm{kp}}^{\frac{1}{\mathrm{n}}}$

According to first condition where pressure is at $100 mm$, it can be written as,

$\frac{10}{1}=k\times {100}^{\frac{1}{n}}$ _________(1)

According to second condition where pressure is at $200 mm$, it can be written as,

$\frac{15}{1}=k\times {200}^{\frac{1}{n}}$ __________(2)

For the Vml of volume be adsorbed at given pressure equal to $300 mm$, it can be written as,

$\frac{V}{1}=k\times {300}^{\frac{1}{n}}$ ___________(3)

Dividing equation (2) by equation (1),

$$\begin{gathered} \frac{15}{10}=2^{\frac{1}{n}} \\ \log \left(\frac{15}{10}\right)=\frac{1}{n}\log 2 \\ \frac{1}{n}=0.585 \end{gathered}$$ 

Dividing equation (3) by equation (1), It can be simplified as,

$$\begin{gathered} \frac{\mathrm{V}}{10}=3^{\frac{1}{\mathrm{n}}} \\ \log \left(\frac{\mathrm{V}}{10}\right)=\frac{1}{\mathrm{n}}\log 3 \end{gathered}$$

$\log \left(\frac{V}{10}\right)=0.585\times 0.4771$

$\frac{V}{10}={10}^{0.279}$

$V={10}^{1.279}$

Compare it with ${10}^x$, we get

$x=1.279\cong 128\times {10}^{-2}$

Hence, the value will be ,$128$.