A 5% by mass of an aqueous solution of A of molar mass $342{\mathrm{gmol}}^{-1}$is isotonic with 0.878% by mass of an aqueous solution of another substance B. The molar mass of B in ${\mathrm{gmol}}^{-1}$ is

Isotonic solution has the same concentration.

C1​=C2​

Where, 

The concentration of one isotonic solution = C1

Concentration of other isotonic solution = C2

$\frac{5}{342}=\frac{0.878}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{B}\left(\mathrm{M}\right)}$

$$\begin{gathered} \Rightarrow 5\times \mathrm{M}=0.878\times 342 \\ \Rightarrow \mathrm{M}=\frac{300.276}{5} \\ \Rightarrow M=60.0552\simeq 60 \end{gathered}$$

Therefore, Option C is the correct answer