$18 \mathrm{g}$ of Glucose is dissolved in $90 \mathrm{g}$ of water. The relative lowering of vapour pressure of solution is

We know that, from Raoult's law for non-volatile solute

$\frac{p°-p_s}{p°}=\frac{n_2}{n_1+n_2}$

 $\frac{p°-p_S}{p°}=\frac{n_2}{n_1+n_2}=\frac{{\displaystyle \frac{w_2}{m_2}}}{{\displaystyle \frac{W_1}{m_1}}+{\displaystyle \frac{w_2}{m_2}}}$

where,

 $w_{2=} mass of solute\left(glucose\right)$  

$m_{2=} molar mass of glucose$

$W_{1=} mass of solvent\left(water\right)$

$m_{1= }molar mass of water$

Given that, $w_2=18g,W_1=90g$

Molar mass of glucose$\left(m_2\right)$=

$\left(C_6{H}_{12}{O}_6\right)=12\times 6+12\times 1+6\times 16=180$

Molar mass of water$\left(m_1\right)$=

$\left({\mathrm{H}}_2\mathrm{O}\right)=2+16=18$

Now, putting the values, we get

$\frac{p°-p_S}{p°}=\frac{{\displaystyle \frac{18}{180}}}{{\displaystyle \frac{90}{18}}+{\displaystyle \frac{18}{180}}}=\frac{{\displaystyle \frac{18}{180}}}{{\displaystyle \frac{900}{180}}+{\displaystyle \frac{18}{180}}}=\frac{18}{918}=\frac{1}{51}$

Therefore, Relative lowering of vapour pressure is $\frac{1}{51}$.

Hence, option(a) is correct.