There is a quick way to solve it!

The largest $6$ -digit number is $\boxed{999600}$ .

The largest $k$ -digit number that is divisible by ${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 },\dots { a }_{ n }$ can be expressed as:

$\left\lfloor \frac { \text{Smallest (k+1) digit number} }{ { \text{ L.C.M of }a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 },\dots { a }_{ n } } \right\rfloor \cdot \,\,\text{ L.C.M of }{a}_{ 1 },{ a}_{ 2 },{ a}_{ 3 },{ a}_{ 4 },{ a}_{ 5 },\dots { a}_{ n}$

Substituting ${a }_{ 1 },{a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }=7,10,15,21,28$ and $k=6$ we have,

$\begin{aligned} & \left\lfloor \frac {\text { Smallest 7 digit number } }{ \text{L.C.M of 7, 10, 15, 21, 28} } \right\rfloor \cdot \, \, \text{L.C.M of 7, 10, 15, 21, 28} \\=&\left\lfloor \frac {\text { 1000000 } }{ 420 } \right\rfloor \cdot \, \, 420\\=&\left\lfloor 2380.952381\right\rfloor\cdot\,\,420\\=&2380\,\,\cdot\,\,420\\=&\boxed{999600}\end{aligned}$