The critical electric field required to produce electron-positron pairs depends on the physical constants $h$, $c$, $m_e$ and $e$. Use dimensional analysis and assume that the dimensionless coefficient is of order one. The magnitude of the critical electric field, in SI units, is of the order

Let the critical electric field be expressed as $E = h^\alpha c^\beta m_e^\gamma e^\delta$.

The dimensional formulas for the given quantities are:
$[E] = M L T^{-3} A^{-1}$
$[h] = M L^2 T^{-1}$
$[c] = L T^{-1}$
$[m_e] = M$
$[e] = A T$

Substituting these dimensions into the assumed relation:
$M L T^{-3} A^{-1} = (M L^2 T^{-1})^\alpha (L T^{-1})^\beta M^\gamma (A T)^\delta$
$M L T^{-3} A^{-1} = M^{\alpha+\gamma} L^{2\alpha+\beta} T^{-\alpha-\beta+\delta} A^\delta$

Equating the powers of the fundamental dimensions on both sides:
For $A$: $\delta = -1$
For $M$: $\alpha + \gamma = 1$
For $L$: $2\alpha + \beta = 1$
For $T$: $-\alpha - \beta + \delta = -3 \Rightarrow -\alpha - \beta - 1 = -3 \Rightarrow \alpha + \beta = 2$

Solving the system of equations:
Subtracting $\alpha + \beta = 2$ from $2\alpha + \beta = 1$, we get $\alpha = -1$.
Substituting $\alpha = -1$ into $\alpha + \beta = 2$, we get $\beta = 3$.
Substituting $\alpha = -1$ into $\alpha + \gamma = 1$, we get $\gamma = 2$.

Thus, the expression for the critical electric field is:
$E = \dfrac{m_e^2 c^3}{e h}$

Substituting the standard SI values of the constants:
$m_e \approx 9.1 \times 10^{-31}$ kg
$c \approx 3 \times 10^8$ m/s
$e \approx 1.6 \times 10^{-19}$ C
$h \approx 6.63 \times 10^{-34}$ J s

$E \approx \dfrac{(9.1 \times 10^{-31})^2 \times (3 \times 10^8)^3}{1.6 \times 10^{-19} \times 6.63 \times 10^{-34}}$

$E \approx \dfrac{82.81 \times 10^{-62} \times 27 \times 10^{24}}{10.6 \times 10^{-53}}$

$E \approx \dfrac{2236 \times 10^{-38}}{10.6 \times 10^{-53}} \approx 211 \times 10^{15} \approx 2.1 \times 10^{17}$ V/m

If we use the reduced Planck constant $\hbar = \dfrac{h}{2\pi} \approx 1.05 \times 10^{-34}$ J s (which is standard for the Schwinger limit in quantum electrodynamics):
$E = \dfrac{m_e^2 c^3}{e \hbar} \approx \dfrac{2236 \times 10^{-38}}{1.6 \times 10^{-19} \times 1.05 \times 10^{-34}} \approx 1.3 \times 10^{18}$ V/m

In either case, the magnitude of the critical electric field is of the order of $10^{18}$ V/m.

Answer: $10^{18}$