Why capacitor value should be greater than gate capacitance of gate of NMOS in Bootstrap topology?
While designing that bootstrap circuit, the value of capacitor is recommended to be 10 time greater than capacitance of gate of transistor.


I didn't know if there's a rule of thumb like this, but the importance comes from the Q (charge), actually.

Here are some technical details:

The total charge held by the bootstrap capacitor is \$Q_B = C_B \cdot V_C\$. When the hi-side MOSFET turns on the total consumed charge is \$Q_T=Q_G + I_G \cdot D\cdot T_{SW}\$; where,


\$Q_G\$ is the gate charge
\$I_G\$ is the gate current (or total current drawn from the capacitor, a few nano-Amps)
\$D\$ is the duty cycle and \$T_{SW}\$ is the switching period (i.e. 1/fSW)


The charge \$Q_T\$ will be provided by \$C_B\$ itself, thus \$Q_B\$ should be sufficiently large. When the charge is consumed, the capacitor voltage should not drop significantly, or else the MOSFET can turn off due to insufficient \$V_{GS}\$. So, \$Q_T\leq C_B\cdot \Delta V_B\$ where \$\Delta V_B\$ is the voltage drop across the bootstrap capacitor.

A practical example:
Using this MOSFET (\$\mathrm{Q_G=18nC}\$) at \$\mathrm{f_{SW}=100kHz}\$ with a duty-cycle of 0.5. Let's assume that the maximum allowed voltage drop is \$\mathrm{\Delta V_B = 100mVDC}\$ and \$\mathrm{I_G=100nA}\$. So, \$\mathrm{C_B\geq (18nC+0.1\mu A \cdot 10\mu s\cdot0.5) \ /\ 0.1V\approx 18nC \ / \ 0.1V = 180nF}\$ which is 50+ times the gate capacitance. If more voltage drop is allowed then the required bootstrap capacitor can be lower. For example, for \$\mathrm{\Delta V_B = 500mVDC}\$ the bootstrap capacitor should be at least 36nF which is 10 times the input capacitance.



SUMMARY

I'm not sure if this can lead to a simplification like "CBOOT should be at least 10 times Ciss of the MOSFET". But sure it depends on the gate charge, actually. If you use buck converter ICs (i.e. the ICs having the MOSFET inside) you should always consider what the datasheet recommends.