1. **State the problem:** Solve for $B$ in the equation $$R=100-\frac{100}{\left(\frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}}+1\right)}.$$ 2. **Rewrite the equation:** Start by isolating the fraction term: $$100 - R = \frac{100}{\left(\frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}}+1\right)}.$$ 3. **Invert both sides:** $$\frac{1}{100-R} = \frac{\frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}}+1}{100}.$$ 4. **Multiply both sides by 100:** $$\frac{100}{100-R} = \frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}} + 1.$$ 5. **Subtract 1 from both sides:** $$\frac{100}{100-R} - 1 = \frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}}.$$ 6. **Simplify the right side fraction:** $$\frac{\frac{E(L-1)+(B-A)}{L}}{\frac{F(L-1)}{L}} = \frac{E(L-1)+(B-A)}{L} \times \frac{L}{F(L-1)} = \frac{E(L-1)+(B-A)}{F(L-1)}.$$ 7. **Set the equation:** $$\frac{100}{100-R} - 1 = \frac{E(L-1)+(B-A)}{F(L-1)}.$$ 8. **Multiply both sides by $F(L-1)$:** $$F(L-1)\left(\frac{100}{100-R} - 1\right) = E(L-1) + (B - A).$$ 9. **Isolate $B$:** $$B = A + F(L-1)\left(\frac{100}{100-R} - 1\right) - E(L-1).$$ **Final answer:** $$\boxed{B = A + F(L-1)\left(\frac{100}{100-R} - 1\right) - E(L-1)}.$$