A triode has a plate resistance of 8 k and a transconductance of...

The plate resistance of a triode is defined as the ratio of change in plate voltage to the change in plate current at a constant grid voltage:

$r_p = \left( \dfrac{\Delta V_p}{\Delta I_p} \right)_{\Delta V_g = 0}$

Rearranging the formula to find the change in plate current:

$\Delta I_p = \dfrac{\Delta V_p}{r_p}$

Given that the increase in plate voltage is $\Delta V_p = 48 \text{ V}$ and the plate resistance is $r_p = 8 \text{ k}\Omega = 8 \times 10^3 \,\Omega$, substituting these values yields:

$\Delta I_p = \dfrac{48}{8 \times 10^3}$

$\Delta I_p = 6 \times 10^{-3} \text{ A} = 6 \text{ mA}$

Since the grid voltage is held constant, the transconductance ($g_m$) is not required for this calculation.