1. Let the infinite nested expression be

$x=\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\cdots}}}$

Then the given quantity is

$6+\log_{3/2}(x).$

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2. Use self-similarity of the infinite radical

Since the same expression repeats inside the square root, we can write

$x=\frac{1}{3\sqrt2}\sqrt{4-x}.$

Now solve for $x$.

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3. Square both sides

$x^2=\frac{1}{18}(4-x)$

because

$\left(\frac{1}{3\sqrt2}\right)^2=\frac{1}{18}.$

So,

$18x^2=4-x$

$18x^2+x-4=0.$

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4. Solve the quadratic equation

$18x^2+x-4=0$

Using the quadratic formula,

$x=\frac{-1\pm\sqrt{1+288}}{36}=\frac{-1\pm17}{36}.$

Thus,

$x=\frac{16}{36}=\frac49 \quad \text{or} \quad x=\frac{-18}{36}=-\frac12.$

Since the nested radical expression is positive, we take

$x=\frac49.$

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5. Evaluate the logarithm

We need

$6+\log_{3/2}\left(\frac49\right).$

Now,

$\frac49=\left(\frac23\right)^2=\left(\frac32\right)^{-2}.$

Hence,

$\log_{3/2}\left(\frac49\right)=\log_{3/2}\left(\left(\frac32\right)^{-2}\right)=-2.$

Therefore,

$6+\log_{3/2}\left(\frac49\right)=6-2=4.$

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6. Final Answer

$\boxed{4}$

The derived answer matches the stored correct answer.