1. Given data

• Two batteries in series, each with emf $E=10\,\text{V}$
• Internal resistances: $r_1=20\,\Omega,\qquad r_2=5\,\Omega$
• External load is a parallel combination of: $30\,\Omega \text{ and } R\,\Omega$

So the total emf of the two batteries in series is $E_{\text{total}}=10+10=20\,\text{V}$ and total internal resistance is $r_{\text{total}}=20+5=25\,\Omega.$

2. Use the condition: voltage across the battery with internal resistance $20\,\Omega$ is zero

For a cell delivering current, terminal voltage is $V = E - Ir$

For the battery with $r=20\,\Omega$, terminal voltage is zero: $0 = 10 - I(20)$

Hence, $I = \frac{10}{20} = 0.5\,\text{A}$

So the current in the series circuit is $I=0.5\,\text{A}.$

3. Find the equivalent resistance of the whole circuit

Using the total series battery combination: $I = \frac{E_{\text{total}}}{r_{\text{total}} + R_{\text{ext}}}$

Substitute values: $0.5 = \frac{20}{25 + R_{\text{ext}}}$

So, $25 + R_{\text{ext}} = \frac{20}{0.5} = 40$

Thus, $R_{\text{ext}} = 40 - 25 = 15\,\Omega$

Therefore the parallel combination of $30\,\Omega$ and $R\,\Omega$ has equivalent resistance $15\,\Omega$.

4. Apply parallel resistance formula

For resistors $30\,\Omega$ and $R\,\Omega$ in parallel: $\frac{30R}{30+R}=15$

Now solve: $30R = 15(30+R)$ $30R = 450 + 15R$ $15R = 450$ $R = 30\,\Omega$

5. Final answer

$\boxed{30}$

So the value of $R$ is $30\,\Omega$.