Given,
Mass of sucrose $\left(w_1\right)=17.1 \mathrm{~g}$
Molar mass of sucrose $\left(M_1\right)=342 \mathrm{~g} \mathrm{~mol}^{-1}$
Mass of oxalic acid $\left(w_2\right)=x g$
Molar mass of oxalic acid $\left(M_2\right)=90 \mathrm{~g} \mathrm{~mol}^{-1}$
Degree of dissociation $(\alpha)$ of oxalic acid $=0.01$
Step I To find the value of van't Hoff factor (i)
$i=(1-\alpha)+n \alpha$
where, $n=$ number of parts in which one molecule of oxalic acid dissociates $(=3)$
$$
\begin{aligned}
& i=(1-0.01)+(3 \times 0.01) \\
& i=0.99+0.03 \\
& i=1.02
\end{aligned}
$$
Step II To find the value of $(x)$
For isotonic solutions :
$\pi$ (sucrose $)=\pi$ (oxalic acid)
$$
[\because \pi=C R T]
$$
or, Molarity (sucrose) = i Molarity (oxalic acid)

$$
\begin{aligned}
& \left.\frac{w_1}{M_1} \text { (sucrose }\right)=i \times\left(\frac{x}{M_2}\right) \text { (oxalic acid) } \\
& x=\frac{w_1}{M_1} \times \frac{M_2}{i}=\frac{171 \times 90}{1.02 \times 342} \\
& x=\frac{1539}{348.84}=4.41 \mathrm{~g}
\end{aligned}
$$
Hence, value of $x=4.41 \mathrm{~g}$
Hence, option (3) is correct.