Let the A.P. have first term $a$ and common difference $d$.

Then $a_n=a+(n-1)d$

Given $a_6=2$: $a+5d=2$ so $a=2-5d$

Now compute the required terms: $a_1=a=2-5d$ $a_4=a+3d=(2-5d)+3d=2-2d$ $a_5=a+4d=(2-5d)+4d=2-d$

Hence the product to be maximised is $P=a_1a_4a_5=(2-5d)(2-2d)(2-d)$

Expand this cubic: First, $(2-2d)(2-d)=4-6d+2d^2$ so $P=(2-5d)(4-6d+2d^2)$ $P=8-12d+4d^2-20d+30d^2-10d^3$ $P=8-32d+34d^2-10d^3$

Differentiate to find critical points: $P'(d)=-32+68d-30d^2$ $P'(d)=0 \\Rightarrow -32+68d-30d^2=0$ $30d^2-68d+32=0$ $15d^2-34d+16=0$

Solve the quadratic: $d=\frac{34\pm\sqrt{34^2-4\cdot 15\cdot 16}}{30}$ $=\frac{34\pm\sqrt{1156-960}}{30}$ $=\frac{34\pm 14}{30}$

Therefore, $d=\frac{48}{30}=\frac{8}{5}$ or $d=\frac{20}{30}=\frac{2}{3}$

Use the second derivative test: $P''(d)=68-60d$

For $d=\frac{8}{5}$: $P''\left(\frac{8}{5}\right)=68-96=-28<0$ so this gives a local maximum.

For $d=\frac{2}{3}$: $P''\left(\frac{2}{3}\right)=68-40=28>0$ so this gives a local minimum.

Hence the common difference that maximises $a_1a_4a_5$ is $\boxed{\frac{8}{5}}$

Comparing with the options, this is Option C.