To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).

Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.

For a monoatomic ideal gas, the molar specific heat at constant volume is:

$$ C_{V,m1} = \frac{3}{2}R $$

For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:

$$ C_{V,m2} = \frac{5}{2}R $$

The weighted average of the molar specific heat for the mixture is:

$$ C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2} $$

Putting the values into the equation, we get:

$$ C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6} $$

$$ C_{V,m} = \frac{(3R + 15R)}{8} $$

$$ C_{V,m} = \frac{18R}{8} $$

$$ C_{V,m} = \frac{9}{4}R $$

Therefore, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$. The correct answer is:

Option D

$$ \frac{9}{4}R $$