At 20$$^\circ$$C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20$$^\circ$$C above an equimolar mixture of benzene and methyl benzene is ___ $$\times 10^{-2}$$. (Nearest integer)

For an equimolar mixture of benzene (B) and methyl benzene (toluene, T), the mole fraction of each in the liquid phase is $$x_B = x_T = 0.5$$.

Using Raoult's law, the partial pressures are: $$P_B = x_B \cdot P_B^* = 0.5 \times 70 = 35 \text{ torr}$$ and $$P_T = x_T \cdot P_T^* = 0.5 \times 20 = 10 \text{ torr}$$.

The total pressure is $$P_{total} = P_B + P_T = 35 + 10 = 45 \text{ torr}$$.

The mole fraction of benzene in the vapour phase is:

$$y_B = \frac{P_B}{P_{total}} = \frac{35}{45} = \frac{7}{9} \approx 0.7778$$

Expressing as $$n \times 10^{-2}$$: $$y_B \approx 77.78 \times 10^{-2}$$, which rounds to $$78 \times 10^{-2}$$.

Therefore, the answer is $$\boxed{78}$$.