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

The plate current is given by:

$i_p = 41(V_p + 7 V_g)^{1.41} \text{ \mu A}$

The plate resistance $r_p$ is defined as the reciprocal of the partial derivative of plate current with respect to plate voltage at a constant grid voltage:

$r_p = \left( \dfrac{\partial V_p}{\partial i_p} \right)_{V_g} = \dfrac{1}{\left( \dfrac{\partial i_p}{\partial V_p} \right)_{V_g}}$

Differentiating $i_p$ with respect to $V_p$ keeping $V_g$ constant:

$\dfrac{\partial i_p}{\partial V_p} = 41 \times 1.41 (V_p + 7 V_g)^{1.41 - 1} \text{ \mu A/V}$

$\dfrac{\partial i_p}{\partial V_p} = 57.81 (V_p + 7 V_g)^{0.41} \text{ \mu A/V}$

Substitute the given operating values $V_p = 250 \text{ V}$ and $V_g = -20 \text{ V}$:

$V_p + 7 V_g = 250 + 7(-20) = 250 - 140 = 110 \text{ V}$

Now, calculate the derivative:

$\dfrac{\partial i_p}{\partial V_p} = 57.81 \times (110)^{0.41} \text{ \mu A/V}$

Using $(110)^{0.41} \approx 6.87$:

$\dfrac{\partial i_p}{\partial V_p} \approx 57.81 \times 6.87 \approx 397.15 \text{ \mu A/V} = 397.15 \times 10^{-6} \text{ A/V}$

The plate resistance is:

$r_p = \dfrac{1}{397.15 \times 10^{-6}} \Omega \approx 2518 \Omega \approx 2.52 \text{ k}\Omega$

The closest matching option is $2.53 \text{ k}\Omega$.

Answer: $2.53 \text{ k}\Omega$