Let $r=|z|$. Since modulus is non-negative, $r\ge 0$.

The inequality becomes $\exp\left(\frac{(r+3)(r-1)}{|r+1|\,\ln 2}\right)\ge \log_{\sqrt2}|5\sqrt7+9i|.$

Simplify $|r+1|$.

Because $r\ge 0$, we have $r+1>0$, so $|r+1|=r+1.$

Hence $\exp\left(\frac{(r+3)(r-1)}{(r+1)\ln 2}\right)\ge \log_{\sqrt2}|5\sqrt7+9i|.$

Compute $|5\sqrt7+9i|$.

$|5\sqrt7+9i|=\sqrt{(5\sqrt7)^2+9^2}=\sqrt{175+81}=\sqrt{256}=16.$

Therefore,

Take natural logarithm on both sides.

Since exponential is increasing, $\frac{(r+3)(r-1)}{(r+1)\ln 2}\ge \ln 8=3\ln 2.$

Multiply by $\ln 2>0$: $\frac{(r+3)(r-1)}{r+1}\ge 3(\ln 2)^2.$

Now check the options to find the least possible value among them satisfying the inequality.

Let $f(r)=\frac{(r+3)(r-1)}{r+1}.$

Since $3(\ln 2)^2\approx 3(0.693)^2\approx 1.44.$

Evaluate options:

• For $r=\sqrt5$:

  =\frac{4+2\sqrt5}{\sqrt5+1}=3\sqrt5-1\over 2?$$ Numerically, $$f(\sqrt5)\approx \frac{(5.236)(1.236)}{3.236}\approx 2.00>1.44.$$ So it satisfies.

• For $r=2$: $f(2)=\frac{(5)(1)}{3}=\frac53>1.44.$ So it satisfies.

• For $r=3$: $f(3)=\frac{(6)(2)}{4}=3>1.44.$ So it satisfies.

• For $r=8$: $f(8)=\frac{11\cdot 7}{9}=\frac{77}{9}>1.44.$ So it satisfies.

Among the given options, the least is $\sqrt5$.

Therefore, the least value of $|z|$ from the options is $\boxed{\sqrt5}.$