1. We need the least value of $a \in \mathbb{R}$ such that $4ax^2+\frac{1}{x}\ge 1 \quad \text{for all } x>0.$

We want this inequality to hold for every positive $x$.

2. Rearrange the inequality: $4ax^2 \ge 1-\frac{1}{x}$ $a \ge \frac{1-\frac{1}{x}}{4x^2}.$

So for all $x>0$, $a$ must be at least $f(x)=\frac{1-\frac{1}{x}}{4x^2}.$

Hence the least possible value of $a$ is $\max_{x>0} f(x).$

3. Simplify $f(x)$: $f(x)=\frac{1}{4x^2}-\frac{1}{4x^3}=\frac{x-1}{4x^3}.$

Let $f(x)=\frac{x-1}{4x^3}.$ We now maximize this for $x>0$.

4. Differentiate: $f(x)=\frac{1}{4}(x-1)x^{-3}.$

Then $f'(x)=\frac{1}{4}\left[x^{-3}-3(x-1)x^{-4}\right].$

Simplify: $f'(x)=\frac{1}{4}x^{-4}\left[x-3(x-1)\right]$ $f'(x)=\frac{1}{4}x^{-4}(x-3x+3)$ $f'(x)=\frac{1}{4}x^{-4}(3-2x).$

5. Critical point: $f'(x)=0 \iff 3-2x=0 \iff x=\frac{3}{2}.$

Since $x^{-4}>0$ for $x>0$, the sign of $f'(x)$ depends on $3-2x$:

• $f'(x)>0$ for $x<\frac{3}{2}$
• $f'(x)<0$ for $x>\frac{3}{2}$

So $f(x)$ attains its maximum at $x=\frac{3}{2}.$

6. Compute the maximum value: $f\left(\frac{3}{2}\right)=\frac{\frac{3}{2}-1}{4\left(\frac{3}{2}\right)^3}$ $=\frac{\frac{1}{2}}{4\cdot \frac{27}{8}}$ $=\frac{\frac{1}{2}}{\frac{27}{2}}$ $=\frac{1}{27}.$

Therefore the least value of $a$ is $a=\frac{1}{27}.$

7. Option check:

• A: $\frac{1}{64}$ — too small
• B: $\frac{1}{32}$ — too small
• C: $\frac{1}{27}$ — correct
• D: $\frac{1}{25}$ — larger than least value

Thus the correct option is $\boxed{\text{C }\frac{1}{27}}.$