Step-by-step Solution:

1. Understand the Geometry of the Area The problem describes the area enclosed by the parabola $y = (1-x)^2$, the x-axis ($y=0$), and the y-axis ($x=0$). The parabola $y = (1-x)^2$ has its vertex at $(1, 0)$ and opens upwards. It intersects the y-axis at $(0, 1)$. The enclosed area is under the curve from $x=0$ to $x=1$.

2. Calculate the Total Area Let $A$ be the total area enclosed by the curves. We can calculate this using a definite integral: $A = \int_{0}^{1} y \,dx = \int_{0}^{1} (1-x)^2 \,dx$ To evaluate the integral, we can use the power rule. Let $u = 1-x$, so $du = -dx$. The limits of integration change from $x=0 \to u=1$ and $x=1 \to u=0$. $A = \int_{1}^{0} u^2 (-du) = \int_{0}^{1} u^2 \,du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$ Alternatively, without substitution: $A = \left[ -\frac{(1-x)^3}{3} \right]_0^1 = \left( -\frac{(1-1)^3}{3} \right) - \left( -\frac{(1-0)^3}{3} \right) = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}$ So, the total area is $\frac{1}{3}$.

3. Define the Sub-regions R₁ and R₂ A vertical line $x=b$ divides the total area into two parts:

   • $R_1$ is the area from $x=0$ to $x=b$: $R_1 = \int_{0}^{b} (1-x)^2 \,dx$
   • $R_2$ is the area from $x=b$ to $x=1$: $R_2 = \int_{b}^{1} (1-x)^2 \,dx$ The sum of these two areas must be the total area: $R_1 + R_2 = A = \frac{1}{3}$.

4. Use the Given Condition to Solve for b We are given the condition $R_1 - R_2 = \frac{1}{4}$. We now have a system of two linear equations for $R_1$ and $R_2$:

   1. $R_1 + R_2 = \frac{1}{3}$
   2. $R_1 - R_2 = \frac{1}{4}$

   Adding the two equations eliminates $R_2$: $(R_1 + R_2) + (R_1 - R_2) = \frac{1}{3} + \frac{1}{4}$ $2R_1 = \frac{4+3}{12} = \frac{7}{12}$ $R_1 = \frac{7}{24}$

5. Calculate R₁ in terms of b and Solve for b Now we evaluate the integral for $R_1$: $R_1 = \int_{0}^{b} (1-x)^2 \,dx = \left[ -\frac{(1-x)^3}{3} \right]_0^b$ $R_1 = \left( -\frac{(1-b)^3}{3} \right) - \left( -\frac{(1-0)^3}{3} \right) = -\frac{(1-b)^3}{3} + \frac{1}{3}$

   We equate this expression for $R_1$ with the value we found in the previous step: $\frac{1 - (1-b)^3}{3} = \frac{7}{24}$ Multiply both sides by 24: $8(1 - (1-b)^3) = 7$ $8 - 8(1-b)^3 = 7$ $8(1-b)^3 = 8 - 7 = 1$ $(1-b)^3 = \frac{1}{8}$ Take the cube root of both sides: $1-b = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$ $b = 1 - \frac{1}{2} = \frac{1}{2}$

6. Conclusion The value of $b$ is $\frac{1}{2}$. This corresponds to option B.