For a 6F6 triode tube, the plate current, plate voltage, and grid...

The mutual conductance $g_m$ of a triode is defined as the rate of change of plate current with respect to grid voltage, keeping the plate voltage constant.

$g_m = \left( \dfrac{\partial i_p}{\partial V_g} \right)_{V_p}$

Given the relation for the plate current:

$i_p = 41(V_p + 7 V_g)^{1.41}$

Differentiating $i_p$ with respect to $V_g$:

$g_m = 41 \times 1.41 \times (V_p + 7 V_g)^{1.41 - 1} \times 7$

$g_m = 404.67 (V_p + 7 V_g)^{0.41}$

Substituting $V_p = 250 \text{ V}$ and $V_g = -20 \text{ V}$:

$g_m = 404.67 (250 + 7(-20))^{0.41}$

$g_m = 404.67 (250 - 140)^{0.41}$

$g_m = 404.67 (110)^{0.41}$

$g_m \approx 404.67 \times 6.87$

$g_m \approx 2780 \mu\text{mho}$

$g_m \approx 2.78 \text{ millimho}$

The closest given option is $2.77 \text{ millimho}$.