Step 1: Identify the colligative property.
Osmotic pressure is asked for, given mass, molar mass, volume and temperature of a dilute non-electrolyte solution.
Step 2: Write the working formula.
Osmotic pressure is $\pi = CRT$, and since molarity $C = \dfrac{n}{V} = \dfrac{W}{M\,V}$, \[ \pi = \frac{W R T}{M\,V}. \]
Step 3: Convert the volume.
$V = 300\ \mathrm{mL} = 0.300\ \mathrm{L}$, with $W = 0.822\ \mathrm{g}$, $M = 340\ \mathrm{g\,mol^{-1}}$, $T = 300\ \mathrm{K}$, $R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$.
Step 4: Compute the moles of solute.
\[ n = \frac{0.822}{340} = 2.418 \times 10^{-3}\ \mathrm{mol}, \quad C = \frac{2.418 \times 10^{-3}}{0.300} = 8.06 \times 10^{-3}\ \mathrm{mol\,L^{-1}}. \]
Step 5: Multiply out $CRT$.
\[ \pi = 8.06 \times 10^{-3} \times 0.0821 \times 300 \approx 0.199\ \mathrm{atm}. \]
Step 6: Round and conclude.
The osmotic pressure is about $0.2\ \mathrm{atm}$, option (B).
\[ \boxed{\pi \approx 0.2\ \mathrm{atm}} \]