1. Parameter form of a focal chord of $y^2=4x$

For the parabola $y^2=4x$, a point with parameter $t$ is $P(t)=(t^2,2t).$

A focal chord has endpoints corresponding to parameters $t_1,t_2$ such that $t_1t_2=-1.$ So we may take $P=(t^2,2t),\qquad Q=\left(\frac{1}{t^2},-\frac{2}{t}\right),$ where the second point corresponds to parameter $-\frac1t$.

2. Use the condition that $\angle P(3,0)Q=\frac\pi2$

Let $R=(3,0)$. Since $\angle PRQ=90^\circ$, we must have $\overrightarrow{RP}\cdot \overrightarrow{RQ}=0.$

Now, $\overrightarrow{RP}=(t^2-3,2t),$ $\overrightarrow{RQ}=\left(\frac{1}{t^2}-3,-\frac{2}{t}\right).$

Thus, \begin{align*} \overrightarrow{RP}\cdot \overrightarrow{RQ} &=(t^2-3)\left(\frac{1}{t^2}-3\right)+(2t)\left(-\frac{2}{t}\right)\ &=(t^2-3)\left(\frac{1}{t^2}-3\right)-4. \end{align*}

Expand: \begin{align*} (t^2-3)\left(\frac{1}{t^2}-3\right) &=1-3t^2-\frac{3}{t^2}+9. \end{align*} So, $1-3t^2-\frac{3}{t^2}+9-4=0,$ which gives $6-3\left(t^2+\frac1{t^2}\right)=0.$ Hence, $t^2+\frac1{t^2}=2.$

Therefore, $\left(t-\frac1t\right)^2=t^2+\frac1{t^2}-2=0,$ so $t=\pm 1.$

Thus the chord endpoints are $P=(1,2),\qquad Q=(1,-2).$ So the line segment $PQ$ is the vertical segment $x=1.$

3. Now use that the same segment is a focal chord of the ellipse

Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a^2>b^2.$ Its foci are at $(\pm c,0),\qquad c^2=a^2-b^2,$ and eccentricity is $e=\frac{c}{a}.$

Since $PQ$ is the same chord with endpoints $(1,2)$ and $(1,-2)$, its midpoint is $(1,0)$. Because this is a focal chord of the ellipse and the chord is vertical, it must pass through a focus. Hence the relevant focus is $(c,0)=(1,0),$ so $c=1.$

Also, since $(1,2)$ lies on the ellipse, $\frac{1}{a^2}+\frac{4}{b^2}=1.$ Using $a^2-b^2=c^2=1,$ let $b^2=m \implies a^2=m+1.$ Then $\frac{1}{m+1}+\frac{4}{m}=1.$ Multiply by $m(m+1)$: $m+4(m+1)=m(m+1).$ So $5m+4=m^2+m,$ which gives $m^2-4m-4=0.$ Hence $m=2+2\sqrt2$ (the positive root only), so $b^2=2+2\sqrt2,$ $a^2=b^2+1=3+2\sqrt2.$

4. Find $\dfrac{1}{e^2}$

Since $e^2=\frac{c^2}{a^2}=\frac{1}{a^2},$ we get $\frac{1}{e^2}=a^2=3+2\sqrt2.$

5. Check options

$\frac{1}{e^2}=3+2\sqrt2,$ which is Option B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.