The voltage gain ($A_v$) of a common emitter (CE) amplifier is given by the relation: $A_v = g_m R_L$ where $g_m$ is the transconductance of the transistor and $R_L$ is the load resistance. When the transistor is replaced in the same amplifier circuit, the load resistance $R_L$ remains constant. Therefore, the voltage gain is directly proportional to the transconductance ($A_v \propto g_m$).

Given: Initial transconductance, $g_{m1} = 0.03 \text{ mho}$ Initial voltage gain, $A_{v1} = G$ New transconductance, $g_{m2} = 0.02 \text{ mho}$

Taking the ratio of the new voltage gain to the initial voltage gain: $\frac{A_{v2}}{A_{v1}} = \frac{g_{m2}}{g_{m1}}$ $\frac{A_{v2}}{G} = \frac{0.02}{0.03} = \frac{2}{3}$ $A_{v2} = \frac{2}{3}G$

Note: The current gain ($\beta$) values provided in the question are extraneous information and are not needed to find the voltage gain when transconductance is given.