Step 1: Tangent to the parabola.
The line $y=mx+c$ touches $y^2=4x$ (so $a=1$) only if $c=\dfrac{a}{m}=\dfrac1m$. So the line is $y=mx+\dfrac1m$.
Step 2: Tangent to the circle.
The same line touches $x^2+y^2=4$ (centre origin, radius $2$) when the perpendicular distance from the origin equals $2$: \[ \frac{|c|}{\sqrt{1+m^2}}=2. \]
Step 3: Substitute c.
With $c=\dfrac1m$, \[ \frac{1/|m|}{\sqrt{1+m^2}}=2. \]
Step 4: Square both sides.
$\dfrac{1}{m^2(1+m^2)}=4$, so $1=4m^2(1+m^2)=4m^2+4m^4$.
Step 5: Solve for m squared.
Let $u=m^2$: $4u^2+4u-1=0$, so $u=\dfrac{-4+\sqrt{16+16}}{8}=\dfrac{-4+4\sqrt2}{8}=\dfrac{\sqrt2-1}{2}$.
Step 6: Form the required quantity.
Then $2m^2=2u=\sqrt2-1$, option (4).
\[ \boxed{2m^2=\sqrt2-1\ \text{(option 4)}} \]