For $NaCl i =2$ so

$\pi_{ NaCl }= i \times C _{ NaCl } \times RT \quad C _{ NaCl }=$ conc. of $NaCl$

$0.1=2 \times C _{ NaCl } \times RT$

$C_{ NaCl }=\frac{0.05}{ RT } \quad C_{\text {glucose }}=$ conc. of glucose

For glucose $i=1$ so

$\pi_{\text {Glucose }}= i \times C _{\text {glucose }} \times RT$

$0.2=1 \times C _{\text {glucose }} \times RT$

$C _{\text {Glucose }}=\frac{0.2}{ RT } \quad \eta_{ NaCl }=$ No. of moles $NaCl$

$\eta_{ NaCl }$ in $1 L = C _{ NaCl } \times V _{\text {Litre }}$

$=\frac{0.05}{ RT } \quad \eta_{ glucose }=$ No. of moles glucose

$\eta_{\text {glucose }}$ in $2 L = C _{\text {glucose }} \times V _{\text {Litre }}$

$=\frac{0.4}{ RT }$

$V _{\text {Total }}=1+2=3 L$

so Final conc. $NaCl =\frac{0.05}{3 RT }$

Final conc. glucose $=\frac{0.4}{3 RT }$

$\pi_{\text {Total }}=\pi_{ NaCl }+\pi_{\text {glucose }}$

$=\left[i \times C_{ NaCl }+C_{\text {glucose }}\right] \times RT$

$=\left(\frac{2 \times 0.05}{3 RT }+\frac{0.4}{3 RT }\right) \times RT$

$=\frac{0.5}{3} atm$

$=0.1666 atm$

$=166.6 \times 10^{-3} atm$

$\Rightarrow 167.00 \times 10^{-3} atm$

so $x=167.00$