Oops, I can't draw an inductor at paint

This is a very long solution

Let at any time the rod has a velocity v and the current flowing be i So we have the equations using faraday's law $\xi =Bvl=L\frac { di }{ dt }$ (i)

$F=mg-Bil=m\frac { dv }{ dt }$ (ii)

Differentiating (ii) w.r.t. time we get $-Bl\frac { di }{ dt } =m\frac { { d }^{ 2 }v }{ d{ t }^{ 2 } }$ (iii)

Using (i) we have $-(\frac { { b }^{ 2 }{ l }^{ 2 } }{ mL } )v=\frac { { d }^{ 2 }v }{ d{ t }^{ 2 } }$ (iv)

This is the equation of SHM. So we write $v=Asin(\omega t+\phi )$ where $\omega =\sqrt { \frac { { b }^{ 2 }{ l }^{ 2 } }{ mL } }$

At $t=0\quad v=0$ $\Rightarrow \phi =0\quad \Rightarrow v=Asin(\omega t)$

Differentiate to get $a=\omega Acos(\omega t)$

$At \quad t=0,a=g\quad \Rightarrow A=\frac { g }{ \omega } \\ \Rightarrow v=\frac { g }{ \omega } sin(\omega t)\quad \quad \Rightarrow x=\frac { g }{ { \omega }^{ 2 } } (1-cos(\omega t))\\ \Rightarrow i=\frac { Bgl }{ { \omega }^{ 2 }L } (1-cos(\omega t))\quad \\ \Rightarrow { v }_{ max }=\frac { g }{ \omega } ,{ x }_{ max }=\frac { 2g }{ { \omega }^{ 2 } } ,{ i }_{ max }=\frac { 2Bgl }{ { \omega }^{ 2 }L }$

Hence we get ${ v }_{ max }{ x }_{ max }{ i }_{ max }=\frac { 4B{ g }^{ 3 }l }{ { \omega }^{ 5 }L }$

$Put\quad the\quad value\quad of \quad \omega \quad to\quad get\quad \\ { v }_{ max }{ x }_{ max }{ i }_{ max }=4{ m }^{ 5/2 }{ g }^{ 3 }{ L }^{ 3/2 }{ B }^{ -4 }{ l }^{ -4 }\\ \Rightarrow a=4,b=5/2,c=3,d=3/2,e=-4,f=-4\\ \Rightarrow \boxed { abcdef=720 }$