(c) Key Idea Firstly, we calculate the osmotic pressure of each solution by using the formula $\pi=C R T$ and then, calculate the total osmotic pressure.
$$
\begin{aligned}
\text { Given, } w_1(\text { urea }) & =1.5 \mathrm{~g} \\
V_1(\text { urea }) & =100 \mathrm{~mL} \\
M_1 & =60 \mathrm{~g} / \mathrm{mol} \\
w_2(\text { cane sugar }) & =3.42 \mathrm{~g} \\
V_2(\text { cane sugar }) & =100 \mathrm{~mL} \\
M_2 & =342 \mathrm{~g} / \mathrm{mol} \\
\pi_1(\text { urea }) & =\frac{w_1 \times 100}{M_1 \times V_1} \times R \times T \\
\pi_1(\text { urea }) & =\frac{1.5 \times 1000}{60 \times 200} \times 0.0821 \times 298 \\
\pi_1(\text { urea }) & =3.05 \mathrm{~atm}
\end{aligned}
$$

Now,
$$
\begin{aligned}
& \pi_2(\text { cane sugar })=\frac{w_2}{M_2 \times V_2} \times 1000 \times R \times T \\
= & \frac{3.42 \times 1000}{342 \times 200} \times 1000 \times 0.0821 \times 298
\end{aligned}
$$

Thus, the osmotic pressure of the resulting solution is
$$
\pi=\pi_1+\pi_2=3.05+1.22=4.27 \mathrm{~atm}
$$

Thus, the nearest value is in accordance with option (c).