Answer
$$8317 \space Pa \space dm^3 \space mol^{-1} \space K^{-1}$$
Work Step by Step
1. Write the perfect gas law, and solve for $R$:
$$pV = nRT \longrightarrow R = \frac{pV}{nT}$$
2. Find p, V, n and T:
$p = p_{H_2O} = \rho gh $
$p = \rho gh = (0.99707 \space g \space cm^{-3} )(9.81 \space ms^{-2})(206.402 \space cm) \times \Big(\frac{100 \space cm}{1 \space m} \Big)^2$
$p = 20188709 \space gm^{-1}s^{-2} \times \frac{1 \space kg}{1000 \space g} \approx 20188.7 \space Pa $
$V = 20.000 \space dm^3$
$$n = \frac{mass}{molar \space mass \space (He)} = \frac{0.25132 \space g}{4.0026 \space g/mol} = 0.062789 \space mol$$
T must be in $K$:
$T(K) = T(C) + 273.15 = 500 + 273.15 = 773.15$
3. Calculate $R$:
$$R = \frac{(20188.7 \space kPa)(20.000 \space dm^3)}{(0.062789 \space mol)(773.15 \space K)} = 8317 \space Pa \space dm^3 \space mol^{-1} \space K^{-1}$$
