A barometer placed inside an upward accelerating elevator reads 76...

Let the upward acceleration of the elevator be $a$.

The effective acceleration due to gravity inside the elevator is $g_{eff} = g + a$.

The air pressure $P$ inside the elevator supports the mercury column in the barometer. Thus, $P = \rho g_{eff} h$, where $\rho$ is the density of mercury and $h$ is the height of the mercury column.

Given that the barometer reads $h = 76 \text{ cm}$, the air pressure is $P = \rho (g + a)(76 \text{ cm})$.

The pressure expressed in terms of standard cm of Hg is $\dfrac{P}{\rho g} = 76 \left(\dfrac{g + a}{g}\right) \text{ cm}$.

Since $a > 0$, we have $\dfrac{g + a}{g} > 1$.

Therefore, the air pressure within the elevator is $> 76 \text{ cm}$ of Hg.