At 300k, 36gm of glucose present per liter in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of solution is 1.52 bar at the same temperature, its concentration  is ----$\times {10}^{-3} {\mathrm{gL}}^{-1}$ (nearest integer)

$\pi =CRT=\frac{W_B\times R\times T}{M_B\times V}$

For both solutions, R, T and V are constants.
For I solution

$(4.98\text{ bar })=\frac{(36 \mathrm{g})\times \mathrm{R}\times \mathrm{T}}{\left(180{\mathrm{gmol}}^{-1}\right)\times \mathrm{V}}$
For II solution

$\text{ (1.52 bar })=\frac{W_B\times R\times T}{M_B\times V}$

On dividing Eq. (ii) by Eq. (i), we get

$\frac{(1.52\text{ bar })}{(4.98\text{ bar })}=\frac{{\mathrm{W}}_{\mathrm{B}}\times \mathrm{R}\times \mathrm{T}}{{\mathrm{M}}_{\mathrm{B}}\times \mathrm{V}}\times \frac{180\times \mathrm{V}}{36\times \mathrm{R}\times \mathrm{T}}$
 

$\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{1.52}{4.98\times 5}=0.0610{ \mathrm{gL}}^{-1}=61\times {10}^{-3}{ \mathrm{gL}}^{-1}$

Hence, the correct answer is 0.0610 g L^-1.