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

Given the equation for the plate current:

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

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

$i_p = 41(250 + 7(-20))^{1.41}$

$i_p = 41(250 - 140)^{1.41}$

$i_p = 41(110)^{1.41}$

Using logarithms or a calculator to evaluate $(110)^{1.41}$:

$\log_{10}(110^{1.41}) = 1.41 \times \log_{10}(110) \approx 1.41 \times 2.0414 \approx 2.878$

$110^{1.41} \approx 10^{2.878} \approx 755.7$

Now, substituting this back into the equation for $i_p$:

$i_p \approx 41 \times 755.7 \mu\text{A}$

$i_p \approx 30984 \mu\text{A}$

Converting microamperes to milliamperes ($1 \text{ mA} = 1000 \mu\text{A}$):

$i_p \approx 30.98 \text{ mA}$

Rounding to the nearest given option, we get $30 \text{ mA}$.